Question

If $y=5^{2\left({\log }_5(x+1)-{\log }_5(3x+1)\right)}$, then what is the value of $\frac{dy}{dx}$ at $x=0$?

Answer 1

$y=5^{2({\log }_5(x+1)-{\log }_5(3x+1))}$

$y=5^{(2{\log }_5(x+1)-2{\log }_5(3x+1))}$

$y=5^{({\log }_5{(x+1)}^2-{\log }_5{(3x+1)}^2)}[\log a^b=b\log a]$

$y=5^{{\log }_5\left(\frac{{(x+1)}^2}{{(3x+1)}^2}\right)}[\log \left(\frac{a}{b}\right)=\log a-\log b]$

$\Rightarrow y=\frac{{(x+1)}^2}{{(3x+1)}^2}$

Differentiating above equation w.r.t. $x$, we have

$\frac{dy}{dx}=\frac{2(x+1){(3x+1)}^2-2(3x+1)\left(3\right){(x+1)}^2}{{(3x+1)}^4}$

$\frac{dy}{dx}=\frac{2(x+1)(3x+1)(3x+1-3x-3)}{{(3x+1)}^4}$

$\Rightarrow \frac{dy}{dx}=\frac{-4(x+1)}{{(3x+1)}^3}$

$\therefore {\frac{dy}{dx}}_{(\text{at}x=0)}=\frac{-4(0+1)}{{((3\times 0)+1)}^3}=-4$

Hence the value of $\frac{dy}{dx}$ at $x=0$ will be $-4$.